enum encoded value is nil while storing the class object in UserDefaults. Codable Protocol is already inherited - swift

Question

I am new to iOS and trying to store User object in UserDefaults. So that when the app is launched again, I can check user type and based on it, I need to navigate to relevant screen.
For that, I have created a User class as below (Codable) and it has one userType enum property!
enum UserType: Int, Codable {
case userType1 = 0
case userType2 = 1
case notDetermined = 2
init(from decoder: Decoder) throws {
let label = try decoder.singleValueContainer().decode(Int.self)
self = UserType(rawValue: label) ?? .notDetermined
}
}
class User: Codable {
public var userFullName: String? = ""
public var userType: UserType? //= .notDetermined
enum CodingKeys: String, CodingKey {
case userFullName
}
}
In my view Controller class, I am creating a new instance for User object and trying to store in user defaults as below:
let newUser = User()
newUser.userFullName = "Test"
newUser.userType = userTypeBtn.isSelected ? .userType1 : .userType2
when I print the newUser's userType, I can see proper value whichever is selected. But after that, when I am trying to store it in userDefaults as below, it returns nil for userType property.
do {
let encoded = try JSONEncoder().encode(newValue)
UserDefaults.standard.set(encoded, forKey: UserDefaultKey.currentUser)
UserDefaults.standard.sync()
} catch {
print("Unable to Encode User Object: (\(error))")
}
when I tried to print this encoded variable, and decoded it in console
JSONDecoder().decode(User.self, from: encoded).userType
it prints nil.
Please help me how can I store optional enum property in UserDefaults and retrieve it when needed using Codable

Answer 1

You should include userType in your CodingKeys enum:
enum CodingKeys: String, CodingKey {
case userFullName
case userType
}
Or just delete the CodingKeys enum entirely, since by default, all the properties are included as coding keys. The keys in the CodingKeys enum determines what the synthesised Codable implementation will encode and decode. If you don't include userType, userType will not be encoded, so it will not be stored into UserDefaults.
I am not getting it from Server and userType is an external property outside the JSON response
This is fine, because userType is optional. If the JSON does not have the key, it will be assigned nil. This might be a problem if you are also encoding User and sending it to the server, and that the server can't handle extra keys in the request, in which case you need two structs - one for storing to/loading from UserDefaults, one for parsing/encoding server response/request.
Remember to encode a new User to UserDefaults when you try this out, since the old one still doesn't have the userType encoded with it.

Answer 2

Observations
Having a custom implementation for Decodable part of enum UserType: Int, Codable is probably not the best idea. Swift compiler supports encoding/decoding enum X: Int out of the box without having you to write custom implementation for it. (In fact, starting with Swift 5.5, Swift compiler can now do this for enums that have cases with associated values as well.)
You should try to avoid having cases like .notDetermined. Either user has a type that's well defined or user.type is nil. You can easily define convenience getters on user itself to know about it's type.
Swift allows nesting of types, so having User.Kind instead of UserType is more natural in Swift.
Following implementation takes care of all of these points.
import Foundation
class User: Codable {
enum Kind: Int, Codable {
case free = 1
case pro = 2
}
public var fullName: String?
public var kind: Kind?
}
let newUser = User()
newUser.fullName = "Test"
newUser.kind = .free
do {
let encoded = try JSONEncoder().encode(newUser)
UserDefaults.standard.set(encoded, forKey: "appUser")
if let fetched = UserDefaults.standard.value(forKey: "appUser") as? Data {
let decoded = try JSONDecoder().decode(User.self, from: fetched)
print(decoded)
}
}
Above code includes definition, construction, encodeAndStore, fetchAndDecode and it does everything you need without any custom implementation.
Bonus
Above code does not print a nice description for the User. For that, you can add CustomStringConvertible conformance like this.
extension User: CustomStringConvertible {
var description: String {
"""
fullName: \(fullName ?? "")
kind: \(kind?.description ?? "")
"""
}
}
extension User.Kind: CustomStringConvertible {
var description: String {
switch self {
case .free: return "free"
case .pro: return "pro"
}
}
}
If you try print(decoded) after implementing this, you will clearly see what you want to see for User instance.
User.kind can be nil and I don't want to handle it with if let every time I need to check this from different screens in the app.
No worries, it can be simplified to this.
extension User {
var isFreeUser: Bool { kind == .free }
var isProUser: Bool { kind == .pro }
}